
A body is orbiting very close to the earth's surface with kinetic energy K.E. The energy required to completely escape from it is:
$\left( A \right)$ $\sqrt 2 KE$
$\left( B \right)$ $2KE$
$\left( C \right)$ $KE\sqrt 2 $
$\left( D \right)$ None of the above
Answer
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Hint: Kinetic energy is the energy of motion, so the object which is moving will have motion and hence it will have kinetic energy. So as energy is of two types that is one is kinetic energy and the other one is potential energy similarly there are different types of kinetic energy like rotational, translational, etc.
Formula used
Kinetic energy,
$KE = \dfrac{1}{2}m{v^2}$; Where $KE$ is the kinetic energy, $m$ is the mass, and $v$ is the velocity.
$PE = mgh$; Where $m$is the mass, $g$ is the acceleration due to gravity, and $h$ will be the height.
Solution
In question, it is given that the body is very close to the earth and so we have to find how much energy the body is required to get completely escaped from it.
When the body is orbiting the earth then the centripetal force is provided by the gravitational force of attraction of earth.
Hence,
$ \Rightarrow \dfrac{{GMm}}{{{R^2}}} = \dfrac{{m{v^2}}}{R}$
Where $G$is the universal gravitational
Kinetic energy will be
$ \Rightarrow KE = \dfrac{{m{v^2}}}{2}$
Which will be equal to $\dfrac{{GMm}}{{2R}}$
Now the gravitational potential energy will be equal to,
\[ \Rightarrow PE = - \dfrac{{GMm}}{R}\]
As we know
After combining both the energy we will get the total energy.
So numerically we can write it as
$ \Rightarrow TE = PE + KE$
Where, $TE$ is the total energy, $PE$ is the potential energy, and $KE$ is the kinetic energy.
And also the total energy is equal to
$ \Rightarrow TE = - \dfrac{{GMm}}{{2R}}$
Therefore $\dfrac{{GMm}}{{2R}}$ which will be equal to $KE$ is required to completely escape from it.
Notes Newton discovered the link between the motion of the Moon and also the motion of a body falling freely on Earth. By his propellant and gravitational theories, he explained Kepler’s laws and established the modern quantitative science of gravitation.
Formula used
Kinetic energy,
$KE = \dfrac{1}{2}m{v^2}$; Where $KE$ is the kinetic energy, $m$ is the mass, and $v$ is the velocity.
$PE = mgh$; Where $m$is the mass, $g$ is the acceleration due to gravity, and $h$ will be the height.
Solution
In question, it is given that the body is very close to the earth and so we have to find how much energy the body is required to get completely escaped from it.
When the body is orbiting the earth then the centripetal force is provided by the gravitational force of attraction of earth.
Hence,
$ \Rightarrow \dfrac{{GMm}}{{{R^2}}} = \dfrac{{m{v^2}}}{R}$
Where $G$is the universal gravitational
Kinetic energy will be
$ \Rightarrow KE = \dfrac{{m{v^2}}}{2}$
Which will be equal to $\dfrac{{GMm}}{{2R}}$
Now the gravitational potential energy will be equal to,
\[ \Rightarrow PE = - \dfrac{{GMm}}{R}\]
As we know
After combining both the energy we will get the total energy.
So numerically we can write it as
$ \Rightarrow TE = PE + KE$
Where, $TE$ is the total energy, $PE$ is the potential energy, and $KE$ is the kinetic energy.
And also the total energy is equal to
$ \Rightarrow TE = - \dfrac{{GMm}}{{2R}}$
Therefore $\dfrac{{GMm}}{{2R}}$ which will be equal to $KE$ is required to completely escape from it.
Notes Newton discovered the link between the motion of the Moon and also the motion of a body falling freely on Earth. By his propellant and gravitational theories, he explained Kepler’s laws and established the modern quantitative science of gravitation.
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